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P2 Past paper exercise watch

1. The point P with x-coordinate p lies on the curve y=x+c\x. Given that the gradient at P is -4 show that c=4p^2.
Any help is appreciated
2. (Original post by toxi)
The point P with x-coordinate p lies on the curve y=x+c\x. Given that the gradient at P is -4 show that c=4p^2.
Any help is appreciated
I did this question before but I can't seem to do it now, is the equation of the curve right? Because the solitary x is mucking me up.
3. y=x + (c/x)

dy/dx= 1 - (c/x^2)

you were told gradient = -4 which = dy/dx
oops
4. my fault, it is y=1+c/x
pls explain me how to do it....I dont wanna copy it i wanna understand it too
5. y = 1 + c/x

dy/dx = - c/x²

=> -4 = - c/p² (by putting x coordinate p in)
=> -4p² = - c
=> c = 4p²
6. (Original post by toxi)
my fault, it is y=1+c/x
pls explain me how to do it....I dont wanna copy it i wanna understand it too
OK:

y = 1+c/x, then dy/dx = -c/x²

At P, dy/dx = -4 and x = p, so substitute.

-4 = -c/p²
4 = c/p²
4p² = c
7. I presume that y = (x+c)/x rather than y = x + (c/x).

Then y = 1 + c/x so dy/dx = -c/x^2.

Since dy/dx = -4 when x = p we have - c/p^2 = -4. So c = 4p^2.
8. lol yes he wrote the question incorrectly
9. got it....pretty easy in the end...thanx
10. (Original post by toxi)
The point P with x-coordinate p lies on the curve y=x+c\x. Given that the gradient at P is -4 show that c=4p^2.
Any help is appreciated
y = (x + c)/x = 1 + cx^-1
dy/dx = -cx^-2 = -c/x^2
-4 = -c/x^2 = -c/p^2
-4p^2 = -c
c = 4p^2

Too late

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